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I was told that it can be shown that $|\sin(x)| \leq 1$ by its definition

$\sin (z):= \frac{1}{2i} \bigl( (\exp(iz)-\exp(-iz)\bigr) $

I am aware that as soon as I choose $x \in \mathbb{R}$ and substitute it into the equation, the equation itself becomes real $\forall x \in \mathbb{R}$

My approach:

$ |\sin (z)| \leq 1 \implies |\sin z|^2 \leq 1 \implies \sin z \cdot \overline{ \sin(z)} \leq 1$ by definition. Since I want to show this for $x \in \mathbb{R}$ it applies that $z= \overline{z}$ and I substitute $x$ into the definition. $$ \frac{1}{2i}((\exp(ix)-\exp(-ix))\cdot \left(-\frac{1}{2i}((\exp(-ix)-\exp(ix))\right) \\ \implies \frac{1}{4}( \exp(0)-\exp(2ix)-\exp(-2ix)\exp(0)) \\ \iff \frac{1}{4}(2-\exp(2ix)-\underbrace{\exp(-2ix)}_{ \in ]0,1]})$$ So this seems to be pretty near to the result I want to proof, however, $\exp(2ix)$ can get sufficiently large for big $x$ so I doubt that my approach is correct.

Any hints, ideas, corrections to keep me going?

Similar Questions

Is it possible to prove $|\sin(x)| \leq 1$, $|\cos(x)| \leq 1$ and $|\sin(x)| \leq |x|$ algebraically?

There's a comment saying that $|\sin(x)| \leq 1$ follows by $ \sin^2(x) + \cos^2(x) =1 $ which I don't see why.

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Note that for non-real $z$, you don't generally have $\lvert\sin z\rvert \leqslant 1$. If you are willing to work with $\lvert e^{ix}\rvert = 1$ for $x\in\mathbb{R}$, it's fairly direct. – Daniel Fischer Nov 16 '13 at 16:20
If you accept $\sin^2 x \cos^2 x = 1$ (which is essentially the Pythagorean theorem), then you get that $\sin ^2 x =1-\cos^2 x \le 1$ and so $|\sin x| \le 1$ – Alfonso Fernandez Nov 16 '13 at 16:22
Thanks for your quick response, so the first line where my implication chain starts is wrong. For $x \in \mathbb{R}$ I don't understand why $| e^{ix} |=1$ if that would be true for $\forall x \in \mathbb{R}$ then it is indeed direct to show. – Spaced Nov 16 '13 at 16:23
For non real $x$ this is simply false, take $2i$ for example.. For real $x$, $\sin^2$ and $\cos^2$ are bigger than zero, so if $\lvert \sin(x)\rvert > 1$ then $\sin^2 + \cos^2 > 1$ a contradiction. – Deven Ware Nov 16 '13 at 16:23
$\lvert e^z\rvert^2 = e^z\overline{e^z} = e^ze^{\overline{z}} = e^{z+\overline{z}} = e^{2\operatorname{Re}z} \Rightarrow \lvert e^z\rvert = e^{\operatorname{Re}z}$ – Daniel Fischer Nov 16 '13 at 16:26

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